Question

1. (10 marks) Calculate the pH after titrating 50.0 mL of a 0.100 M weak base solution (K_b = 7 x10^-9) with:
   1. 0.200 M HBr to the equivalence point

b. 50.0 mL of 0.200 M HBr

Answer 1

a)

find the volume of HBr used to reach equivalence point

M(B)*V(B) =M(HBr)*V(HBr)

0.1 M *50.0 mL = 0.2M *V(HBr)

V(HBr) = 25 mL

Given:

M(HBr) = 0.2 M

V(HBr) = 25 mL

M(B) = 0.1 M

V(B) = 50 mL

mol(HBr) = M(HBr) * V(HBr)

mol(HBr) = 0.2 M * 25 mL = 5 mmol

mol(B) = M(B) * V(B)

mol(B) = 0.1 M * 50 mL = 5 mmol

We have:

mol(HBr) = 5 mmol

mol(B) = 5 mmol

5 mmol of both will react to form BH+ and H2O

BH+ here is strong acid

BH+ formed = 5 mmol

Volume of Solution = 25 + 50 = 75 mL

Ka of BH+ = Kw/Kb = 1.0E-14/7.0E-9 = 1.429*10^-6

concentration ofBH+,c = 5 mmol/75 mL = 0.0667 M

BH+ + H2O -----> B + H+

6.667*10^-2 0 0

6.667*10^-2-x x x

Ka = [H+][B]/[BH+]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((1.429*10^-6)*6.667*10^-2) = 3.086*10^-4

since c is much greater than x, our assumption is correct

so, x = 3.086*10^-4 M

[H+] = x = 3.086*10^-4 M

use:

pH = -log [H+]

= -log (3.086*10^-4)

= 3.5106

Answer: 3.51

b)

Given:

M(HBr) = 0.2 M

V(HBr) = 50 mL

M(B) = 0.1 M

V(B) = 50 mL

mol(HBr) = M(HBr) * V(HBr)

mol(HBr) = 0.2 M * 50 mL = 10 mmol

mol(B) = M(B) * V(B)

mol(B) = 0.1 M * 50 mL = 5 mmol

We have:

mol(HBr) = 10 mmol

mol(B) = 5 mmol

5 mmol of both will react

excess HBr remaining = 5 mmol

Volume of Solution = 50 + 50 = 100 mL

[H+] = 5 mmol/100 mL = 0.05 M

use:

pH = -log [H+]

= -log (5*10^-2)

= 1.301

Answer: 1.30